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The Real And Complex Number Systems

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Since f a f p 0, we have f q 1 0whereq 1 a, p and since<br />

f p f b 0, we have f q 2 0whereq 2 p, b by Rolle’s <strong>The</strong>orem. Since<br />

f q 1 f q 2 0, we have f c 0wherec q 1 , q 2 by Rolle’s <strong>The</strong>orem.<br />

5.22 Assume f has a finite derivative in some interval a, .<br />

(a) If fx 1andf x c as x , prove that c 0.<br />

Proof: Consider fx 1 fx f y where y x, x 1 by Mean Value <strong>The</strong>orem,<br />

since<br />

lim fx 1<br />

x<br />

which implies that<br />

limfx 1 fx 0<br />

x<br />

which implies that (x y )<br />

lim<br />

x f y 0 y<br />

lim fy<br />

Since f x c as x , we know that c 0.<br />

Remark: (i) <strong>The</strong>re is a similar exercise; we write it as follows. If fx L and<br />

f x c as x , prove that c 0.<br />

Proof: By the same method metioned in (a), we complete it.<br />

(ii) <strong>The</strong> exercise tells that the function is smooth; its first derivative is smooth too.<br />

(b) If f x 1asx , prove that fx/x 1asx .<br />

Proof: Given 0, we want to find M 0 such that as x M<br />

fx<br />

x 1 .<br />

Since f x 1asx , thengiven 3 ,thereisM 0 such that as x M ,we<br />

have<br />

|f x 1| <br />

3 |f x| 1 <br />

3<br />

<br />

By Taylor <strong>The</strong>orem with Remainder Term,<br />

fx fM f x M <br />

fx x fM f 1x f M ,<br />

then for x M ,<br />

fx<br />

x 1 fM <br />

x |f 1| f M <br />

x<br />

<br />

fM <br />

x 3 1 3<br />

Choose M 0 such that as x M M ,wehave<br />

fM<br />

x <br />

3 M<br />

and x<br />

<br />

M <br />

x by (*)<br />

*<br />

**<br />

/3<br />

1 3 . ***<br />

Combine (**) with (***), we have proved that given 0, there is a M 0 such that as<br />

x M, wehave<br />

fx<br />

x 1 .<br />

That is, lim x<br />

fx<br />

x 1.

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